Question

Show that any positive odd integer is of the form $6q + 1$, or $6q + 3$, or $6q + 5$, where $q$ is some integer.

Answer 1

Here, we have to prove that any positive odd integer is of the given form. We will use Euclid’s lemma to prove that the given form is of the integer form. Then by substituting the odd integers for the remainder, we will prove that it is of the form of a positive odd integer. Odd integers are the integers which cannot be exactly divided into pairs.

Formula Used:
Euclid’s Lemma: $a = bq + r$, where $a,b$ are two positive integers, $q$ is the quotient and $r$ is the remainder such that $0 \le r < b$.

Complete Step by Step Solution:
We are given to prove that any positive odd integer is of the form $6q + 1$, or $6q + 3$, or $6q + 5$, where $q$ is some integer.
Let $a$ be the positive integer and $b = 6 \in {\bf{Z}}$. Thus by using the Euclid’s Lemma $a = bq + r$, we get
$a = 6q + r$ ……………………………………………………………..$\left( 1 \right)$
Here, $0 \le r < 6$, so, $r$ is an integer which is less than or equal to 6 and has the value of $r$ as either 0, 1, 2, 3, 4, or 5.
Thus $a = 6q + r$ is an integer.
We know that when an integer is multiplied with an even number, then it would be a positive even integer.
Thus, $6q$ is an even integer.
Now, we will prove that any positive odd integer is of the form $6q + 1$, or $6q + 3$, or $6q + 5$
Now, by substituting $r = 1$ in the equation $\left( 1 \right)$, we get
$a = 6q + 1$
We know that when an even integer is added with an odd integer, then the resulting integer would be a positive odd integer.
Thus, the integer will always be a positive odd integer.
Now, by substituting $r = 3$ in the equation $\left( 1 \right)$, we get
$a = 6q + 3$
We know that when an even integer is added with an odd integer, then the resulting integer would be a positive odd integer.
Thus, the integer will always be a positive odd integer.
Now, by substituting $r = 5$ in the equation $\left( 1 \right)$, we get
$a = 6q + 5$
We know that when an even integer is added with an odd integer, then the resulting integer would be a positive odd integer.
Thus, the integer will always be a positive odd integer.
Thus, we proved that any positive odd integer is of the form $6q + 1$, or $6q + 3$, or $6q + 5$.

Therefore, any positive odd integer is of the form $6q + 1$, or $6q + 3$, or $6q + 5$, where $q$ is some integer.

Note:
We should know that Euclid’s lemma states that if we have two positive integers $a$ and $b$, then there exist unique integers $q$ and $r$ which satisfies the condition $a = bq + r$ where $0 \le r < b$. Euclid’s lemma is the basis for Euclid’s division algorithm. We can also remember it in such a way that $a$ is the dividend, $b$ is the divisor, $q$ is the quotient and $r$ is the remainder. We know that remainder should always be less than the divisor. We should also remember the properties of odd integers.